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04242012, 09:57 AM  #1 
Bug Crusher

Drawing the Vector Diagram for DV Tool
Hello,
I'm trying to draw the vector diagram for a plane change + orbital circularization maneuver to get to GEO. I want to try to calculate the components by hand. This is the setup in IMFD using the DV tool: I want to plane change (RIn) 24 degrees. My apogee arrival (ApV) velocity is 1605 m/s. My required circularization velocity (Cir) is 1477. When I try to set up the vectors, I get this : (green is current orbit, red is burn vector, blue is target orbit, black is what I think the components are) When I calculate the plane change component (dVp), as shown, using the law of cosines, I get about 1282 m/s. However, as seen in the first post, the actual plane change component magnitude is much less, in the neighborhood of 950 m/s. I'm wondering if this 25% discrepancy has to do with the "vel. frame" (how the frame is being considered). I didn't completely understand this in the manual, or in the FULL manual, which I have also read. Any input would be appreciated. 
04242012, 11:59 PM  #2 
Orbinaut

I'm not strong in Math skills, so I'm not sure exactly where your calculations are going wrong. I can offer some advice on using IMFD, and some observations based on the screenshots above.
Velocity Frame is appropriate for this manuever. In Vel. Frame, the ship's velocity vector is the "reference". dVf is "forward", or prograde (at the time of the burn). dVi, or "inward", is at right angles to dVf, and generally toward the center of the reference body. dVp, or "planar", is at right angles to both dVf and dVi, and positive dVp is in the "normal" direction. Note that this means that one of the angles on your triangle (I believe angle C) will ALWAYS be 90 degrees. One thing I noticed is that you aren't actually getting into GEO with this burn  your PeA doesn't match your ApA by a significant amount. You'll need more dVf  and since the dV required for a plane change is dependant on the velocity of the ship, you'll need more dVp as well. Also, there is a 800 second difference between your TEj and your ApT  you have planned the burn too early (TEj should be ApT minus about 1/3 the burntime). The data displayed by Delta Velocity program isn't that accurate  you should be using Map (with Plan enabled  course will be shown in blue, not green) for more accuracy. A better plan would be to separate the plane change from the circularization burn. Plan the initial transfer to GEO so that the plane change is just before the circularization burn (your velocity will be at it's lowest), then center the circularization burn around the Apo. It looks like the transfer burn wasn't centered on the node correctly (should have had 1/3 of the burn before the node, and 2/3 after)  this is why you need to have the planned burn start a bit early for circularization (because you need to also center it on the node  and the node and the Apo aren't in quite the same place. That could also be an effect of precession if you have "nonspherical gravity sources" enabled in the launchpad. Note that I say to center the burns by "thirds", not "halfs". This is to account for the burn integration used by IMFD  it's not perfect but a bit more accurate than "halfs". If you are determined to use a single burn for both plane change and circularization, you'll need to be more accurate on the transfer burn. 
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04252012, 02:43 AM  #3 
Bug Crusher

Thank you for your response, however, performing the maneuver is not the issue.
First of all, I'm sure I want to do a combined burn: http://www.braeunig.us/space/orbmech.htm#maneuver (see section orbital plane changes) Very best would be a bielliptic transfer from supersynchronous orbit, but I don't want to do that. I intentionally tweaked the burn start time a little bit to improve the amount of plane change and to get ApA closer to 39786km. From here, it takes a less than 8 m/s Hohmann transfer to get to a pretty good GEO, so I don't think these errors are too significant. My main interests are calculating the vectors by hand and drawing the diagram. To calculate the segment I have labeled "dVp", I use the law of cosines, which simplifies to: 2 x (ApV + Cir) x SIN(RIn/2) = 2x(1605+1477)xSIN(24/2) = 1282 m/s This isn't even close to what you have to enter into IMFD as the "dVp" component (948 ). The discrepancy on the order of hundreds of m/s, the best I can tell, can't be caused simply be due to starting the burn a few minutes early, etc. Actually, when I perform the burn as shown, I end up with a final velocity of 3082 m/s, which is exactly 1477+1605. So what I have labeled as "dVp" must be 1282 m/s. RIn is about 0.04, which only needs about 2 m/s to get to as close to 0 as I can get. So, the issue seems to be that this is not what IMFD considers to be dVp (Unless my math is wrong). What vector do I draw to represent dVp in IMFD? I have a feeling that this is where Vel. Frame is coming into play. You are rotating during the burn. So is the "plane change" direction constantly being redefined during the maneuver? Also, the vector that I have labeled "dVp" is not perpendicular to dVf. 
04252012, 03:58 AM  #4 
Orbinaut

What was the starting orbit? If it was circular, then the timing of the burn won't have anything to do with the ApA  that will be determined by the length of the burn. "Adjusting" the TEj of the transfer burn was a mistake  for this to work you need to center both the transfer burn and the circ burn on the nodes. That is true of ALL offplane transfers. If you look, your current ApA is 35.58k, and the ApA after the burn would be 35.77. This difference is due to the burn NOT being properly centered on the Apo.
As to the dVp vector, IMFD calculates all burns as point impulse burns (ie, dV is instant) so the vector will be at a right angle to the prograde vector at TEj=0. Therefore, there is some small inaccuracy (since the burn takes time), but the vectors don't "rotate" during the burn. I did notice that your vectors weren't at right angles  but was unsure that was intentional. dVf and dVp MUST be at right angles. Perhaps this is where the error is. By having the angle between dVf and dVp other than 90 degrees, you end up with an incorrect "burn angle", which will throw off all calculations and the "red line" will be wrong. In fact, if I had the time, I'd trig this out using basic "right triangle" trigonometry  not the law of cosines (which is only needed for nonright triangles). dVf will be one leg, the "side adjacent", the dVp will be the "side opposite" and the burn vector will be the hypotenuse. The pythagorean theorem can be used to determine the length of the hypotenuse, then dVp / dVf = tan(burnangle). Or dVp / burnvector = sine(burnangle), however you want to do it. I'm not sure it matters, but your diagram seems to be using "equatorial frame" rather than "velocity frame"  but I don't think that makes any real difference here. You'll need someone with much better mathfu to help with that. I'm thinking that this diagram is a bit more accurate  the vertical axis is the equator. The angle between the green line and the purple line is the RInc, and the burn angle = RInc * 2. I could be wrong. untitled.jpg also, ignore the typo  dFv in the diagram should be dVf. Last edited by Tommy; 04252012 at 04:16 AM. 

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